choirgirl1987 's question at Yahoo! Answers (Jordan block)

[Fernando Revilla]

Here is the question:

I'm just not sure what -J(lambda) looks like as a matrix. Don't need the full answer, just what would -J(lambda) be...

Here is a link to the question:

What is the Jordan Canonical form of -J(lambda)? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

 

[Fernando Revilla]

Hello choirgirl1987,

A Jordan block $J(\lambda)$ is a matrix of the form:

$$J(\lambda)=\begin{bmatrix} \lambda & 1 & 0 &\ldots & 0 & 0 & 0\\ 0 & \lambda & 1 &\ldots & 0&0&0 \\0 & 0 & \lambda &\ldots & 0&0&0 \\\vdots&&&&&&\vdots \\ 0 &0 & 0 &\ldots & \lambda & 1&0\\0 &0 &0 &\ldots &0&\lambda & 1\\0 & 0 &0&\ldots & 0&0&\lambda\end{bmatrix}$$ and a Jordan normal form is a block diagonal matrix of de form $$J=\begin{bmatrix} J(\lambda_1) & 0 & \ldots & 0\\ 0 & J(\lambda_2) & \ldots & 0 \\ \vdots&&&\vdots \\ 0 & 0 &\ldots & J(\lambda_p)\end{bmatrix}$$If you have further questions, you can post them in the Linear and Abstract Algebra section.